NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev

Mathematics (Maths) Class 6

Created by: Praveen Kumar

Class 6 : NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev

The document NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev is a part of the Class 6 Course Mathematics (Maths) Class 6.
All you need of Class 6 at this link: Class 6

Exercise 14.4
Question 1: Draw any line segment NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev. Mark any point M on it. Through M, draw a perpendicular to NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev. (Use ruler and compasses)
Answer 1:
Steps of construction:
(i) Draw a line segmentNCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev and mark a point M on it.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
(ii) Taking M as centre and a convenient radius, construct an arc intersecting the line segment NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev at points C and D respectively.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(iii) By taking centres as C and D and radius greater than CM, construct two arcs such that they intersect each other at point E.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
(iv) Join EM. Now NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRevis perpendicular toNCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev 
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
Question 2: Draw any line segment NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev. Take any point R not on it. Through R, draw a perpendicular to NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev. (Use ruler and set-square)
Answer 2:
(i) Draw a given line segmentNCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev and mark a point R outside the line segment NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
(ii) Place a set square on NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev such that one of its right angles arm aligns along NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(iii) Now, place the ruler along the edge opposite to right angle of set square.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(iv) Hold the ruler fixed. Slide the set square along the ruler such that the point R touches the other arm of set square.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(v) Draw a line along this edge of set square which passes through point R. Now, it is the required line perpendicular toNCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev


Question 3:  Draw a line l and a point X on it. Through X, draw a line segment NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev perpendicular to l. Now draw a perpendicular to NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev to Y. (use ruler and compasses) 
Answer 3: 
Steps of construction:
(i) Draw a line l and mark a point X on it.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(ii) By taking X as centre and with a convenient radius, draw an arc intersecting the line l at points A and B respectively.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(iii) With A and B as centres and a radius more than AX, construct two arcs such that they intersect each other at point Y.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(iv) Join XY. Here NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev is perpendicular to l
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRevSimilarly, by taking C and D as centres and radius more than CY, construct two arcs intersecting at point Z. Join ZY. The line NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev is perpendicular to NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev at Y
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
Exercise 14.5
Question 1: Draw NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev of length 7.3 cm and find its axis of symmetry.
Answer 1: Axis of symmetry of line segment NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev will be the perpendicular bisector of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev. So, draw the perpendicular bisector of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
Steps of construction:
(i) Draw a line segment NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev = 7.3 cm
(ii) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.
(iii) Join CD. Then CD is the axis of symmetry of the line segment AB.

Question 2:
Draw a line segment of length 9.5 cm and construct its perpendicular bisector.
Answer 2:
Steps of construction:
(i) Draw a line segment NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev = 9.5 cm
(ii) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.
(iii) Join CD. Then CD is the perpendicular bisector of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev .

NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev


Question 3: Draw the perpendicular bisector of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev whose length is 10.3 cm.
(a) Take any point P on the bisector drawn.
Examine whether PX = PY.
(b) If M is the mid-point of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev, what can you say about the lengths MX and XY?
Answer 3:
Steps of construction:

NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRevNCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(i) Draw a line segment NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev = 10.3 cm
(ii) Taking X and Y as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.
(iii) Join CD. Then CD is the required perpendicular bisector of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev .
Now:
(a) Take any point P on the bisector drawn. With the help of divider we can check that NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
(b) If M is the mid-point of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev


Question 4: 
Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.
Answer 4:
Steps of construction:
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(i) Draw a line segment AB = 12.8 cm
(ii) Draw the perpendicular bisector of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev which cuts it at C. Thus, C is the midpoint of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev.
(iii) Draw the perpendicular bisector of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev which cuts it at D. Thus D is the midpoint of.
(iv) Again, draw the perpendicular bisector of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev which cuts it at E. Thus, E is the mid-point of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev.
(v) Now, point C, D and E divide the line segment NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev in the four equal parts.
(vi) By actual measurement, we find that NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev


Question 5:
With NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev of length 6.1 cm as diameter, draw a circle.
Answer 5:
Steps of construction:
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
(i) Draw a line segment NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev = 6.1 cm.
(ii) Draw the perpendicular bisector of PQ which cuts, it at O. Thus O is the mid-point of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev .
(iii) Taking O as centre and OP or OQ as radius draw a circle where diameter is the line segment NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev .


Question 6:
Draw a circle with centre C and radius 3.4 cm. Draw any chord NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev. Construct the perpendicular bisector NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev and examine if it passes through C.
Answer 6:
Steps of construction:
(i) Mark any point C on the sheet
(ii) Adjust the compasses up to 3.4 cm and by putting the pointer of compasses at point C, turn compasses slowly to draw the circle. This is the required circle of 3.4 cm radius.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev

(iii) Mark any chord NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev in the circle
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(iv) Now, taking A and B as centres, draw arcs on both sides of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev. Let these intersect each other at points D and E.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(v) Join DE. Now DE is the perpendicular bisector of AB.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
If NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev is extended, it will pass through point C.

Question 7: Repeat Question 6, if NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev happens to be a diameter.
Answer 7:
Steps of construction:
(i) Mark any point C on the sheet.
(ii) Adjust the compasses up to 3.4 cm and by putting the pointer of compasses at point C, Turn the compasses slowly to draw the circle. This is the required circle of 3.4 cm
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(iii) Now mark any diameter NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev in the circle.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(iv) Now taking A and B as centres, draw arcs on both sides of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev with radius more than NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev.  Let these intersect each other at points D and E.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(v) Join DE, which is perpendicular bisector of AB.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
Now, we may observe that NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev is passing through the centre C of the circle.


Question 8:
Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?
Answer 8:
Steps of construction:
(i) Mark any point C on the sheet. Now adjust the compasses up to 4 cm and by placing the pointer of compasses at point C, turn the compasses slowly to draw the circle. This is the required circle of 4 cm radius
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(ii) Take any two chords NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev in the circle
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(iii) By taking A and B as centres and radius more than half of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev draw arcs on both sides of AB. The arcs are intersecting each other at point E and F. Join EF which is perpendicular bisector of AB.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(iv) Again take C and D as centres and radius more than half of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev draw arcs on both sides of CD such that they are intersecting each other at points G, H. Join GH which is perpendicular bisector of CD
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev

We may observe that when EF and GH are extended they meet at the point O, which is the centre of circle

Question 9:
Draw any angle with vertex O. Take a point A on one of its arms and B on another such that OA = OB. Draw the perpendicular bisectors of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev and NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev. Let them meet at P. Is PA = PB?
Answer 9:
Steps of construction:
(i) Draw any angle with vertex as O.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(ii) By taking O as centre and with convenient radius, draw arcs on both rays of this angle. Let these points are A and B
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(iii) Now take O and A as centres and with radius more than half of OA, draw arcs on both sides of OA. Let these intersects at points C and D respectively. Join CD
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(iv) Similarly we may find NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev which is perpendicular bisector of NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev. These perpendicular bisectors NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev intersects each other at point P. Now measure PA and PB. They are equal in length.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev


Exercise 14.6
Question 1:
Draw ∠ POQ of measure 75° and find its line of symmetry.
Answer 1:
Steps of construction:
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(i) Draw a line l and mark a point O on it.
(ii) Place the pointer of the compasses at O and draw an arc of any radius which intersects the line l at A.
(iii) Taking same radius, with centre A, cut the previous arc at B.
(iv) Join OB, then ∠ BOA = 60 .
(v) Taking same radius, with centre B, cut the previous arc at C.
(vi) Draw bisector of ∠BOC. The angle is of 90 . Mark it at D. Thus, ∠DOA = 90o 
(vii) Draw NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev as bisector of ∠DOB. Thus, ∠POA = 75o.


Question 2: Draw an angle of measure 147o  and construct its bisector.
Answer 2:
Steps of construction:
Following steps are followed to construct an angle of measure 1470 and its bisector
(i) Draw a line l and mark point O on it. Place the centre of protractor at point O and the zero edge along line l
(ii) Mark a point A at an angle of measure 1470. Join OA. Now OA is the required ray making 1470 with line l
(iii) By taking point O as centre, draw an arc of convenient radius. Let this intersect both rays of angle 1470 at points A and B.
(iv) By taking A and B as centres draw arcs of radius more than 1 / 2 AB in the interior angle of 1470. Let these intersect each other at point C. Join OC.
OC is the required bisector of 1470 angle
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev


Question 3:
Draw a right angle and construct its bisector.
Answer 3:
Steps of construction:
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(i) Draw a line PQ and take a point O on it.
(ii) Taking O as centre and convenient radius, draw an arc which intersects PQ at A and B.
(iii) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C.
(iv) Join OC. Thus, ∠COQ is the required right angle.
(v) Taking B and E as centre and radius more than half of BE, draw two arcs which intersect each other at the point D.
(vi) Join OD. Thus, NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev is the required bisector of ∠COQ.


Question 4:
Draw an angle of measure 153o and divide it into four equal parts.
Answer 4:
Steps of construction:
Following steps are followed to construct an angle of measure 1530 and its bisector
(i) Draw a line l and mark a point O on it. Place the centre of protractor at point O and the zero edge along line l
(ii) Mark a point A at the measure of angle 1530. Join OA. Now OA is the required ray making 1530 with line l
(iii) Draw an arc of convenient radius by taking point O as centre. Let this intersects both rays of angle 1530 at points A and B.
(iv) Take A and B as centres and draw arcs of radius more than 1 / 2 AB in the interior of angle of 1530. Let these intersect each other at C. Join OC
(v) Let OC intersect major arc at point D. Draw arcs of radius more than 1 / 2 AD with A and D as centres and also D and B as centres. Let these are intersecting each other at points E and F respectively. Now join OE and OF
OF, OC, OE are the rays dividing 1530 angle into four equal parts.
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev

Question 5:
Construct with ruler and compasses, angles of following measures: 
(a) 60° 
(b) 30° 
(c) 90° 
(d) 120° 
(e) 45° 
(f) 135°
Answer 5:
Steps of construction:
(a) 60°
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(i) Draw a ray NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev .
(ii) Taking O as centre and convenient radius, mark an arc, which intersects NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev at P.
(iii) Taking P as centre and same radius, cut previous arc at Q.
(iv) Join OQ.
Thus, ∠BOA is required angle of 60° .

(b) 30o

NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(i) Draw a ray NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev .
(ii) Taking O as centre and convenient radius, mark an arc, which intersects NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev at P.
(iii) Taking P as centre and same radius, cut previous arc at Q.
(iv) Join OQ. Thus, ∠BOA is required angle of 60o.
(v) Put the pointer on P and mark an arc.
(vi) Put the pointer on Q and with same radius, cut the previous arc at C. Thus, ∠COA is required angle of 30o.

(c) 90°
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(i) Draw a ray NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev.
(ii) Taking O as centre and convenient radius, mark an arc, which intersects NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev at X.
(iii) Taking X as centre and same radius, cut previous arc at Y.
(iv) Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.
(v) Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.
(vi) Join OS and produce it to form a ray OB.
Thus, ∠BOA is required angle of 900.

(d) 120°
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(i) Draw a ray NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev.
(ii) Taking O as centre and convenient radius, mark an arc, which intersects NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev at P.
(iii) Taking P as centre and same radius, cut previous arc at Q.
(iv) Taking Q as centre and same radius cut the arc at S.
(v) Join OS.
Thus, ∠AOD is required angle of 120°.

(e) 45o 
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(i) Draw a ray NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev .
(ii) Taking O as centre and convenient radius, mark an arc, which intersects NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev at X.
(iii) Taking X as centre and same radius, cut previous arc at Y.
(iv) Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.
(v) Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.
(vi) Join OS and produce it to form a ray OB. Thus, ∠BOA is required angle of 90°.
(vii) Draw the bisector of ∠BOA.
Thus, ∠MOA is required angle of 45°.

(f) 135°

NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev

(i) Draw a line PQ and take a point O on it.
(ii) Taking O as centre and convenient radius, mark an arc, which intersects PQ at A and B.
(iii) Taking A and B as centres and radius more than half of AB, draw two arcs intersecting each other at R.
(iv) Join OR. Thus, ∠QOR = ∠POQ = 90°.
(v) Draw NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev the bisector of ∠POR.
Thus, ∠QOD is required angle of 135°.

Question 6:
Draw an angle of measure 45o and bisect it.
Answer 6:
Steps of construction:
Following steps are followed to construct an angle of measure 450 and its bisector.
(i) Using the protractor ∠POQ of 450 measure may be formed on a line l
(ii) Draw an arc of convenient radius with centre as O. Let this intersects both rays of angle 450 at points A and B
(iii) Take A and B as centres, draw arcs of radius more than 1 / 2 AB in the interior of angle of 450. Let these intersect each other at C. Join OC
OC is the required bisector of 450 angle
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
Question 7:
Draw an angle of measure 135o and bisect it.
Answer 7:
Steps of construction:
Following steps are followed to construct an angle of measure 1350 and its bisector.
(i) By using a protractor ∠POQ of 1350 measure may be formed on a line l
(ii) Draw an arc of convenient radius by taking O as centre. Let this intersect both rays of angle 1350 at points A and B respectively.
(iii) Take A and B as centres, draw arcs of radius more than 1 / 2 AB in the interior of angle of 1350.
Let these intersect each other at C. Join OC.
OC is the required bisector of 1350 angle
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
Question 8:
Draw an angle of 70o. Make a copy of it using only a straight edge and compasses.
Answer 8:
Steps of construction:
(i) Draw an angle 70o with protractor, i.e., ∠POQ = 70o
(ii) Draw a ray NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev
(iii) Place the compasses at O and draw an arc to cut the rays of ∠POQ at L and M.
(iv) Use the same compasses, setting to draw an arc with A as centre, cutting AB at X.
(v) Set your compasses setting to the length LM with the same radius.
(vi) Place the compasses pointer at X and draw the arc to cut the arc drawn earlier at Y.
(vii) Join AY. Thus, ∠YAX = 70o

Question 9:
Draw an angle of 40o. Copy its supplementary angle.
Answer 9:
Steps of construction:
NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRevNCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev(i) Draw an angle of 40° with the help of protractor, naming ∠AOB.
(ii) Draw a line PQ.
(iii) Take any point M on PQ.
(iv) Place the compasses at O and draw an arc to cut the rays of ∠AOB at L and N.
(v) Use the same compasses setting to draw an arc O as centre, cutting MQ at X.
(vi) Set your compasses to length LN with the same radius.
(vii) Place the compasses at X and draw the arc to cut the arc drawn earlier Y.
(viii) Join MY.
Thus, ∠QMY = 40° and ∠PMY is supplementary of it.

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